The New Sudoku Players' Forum

[daj95376]

I've withdrawn my example and subsequent related posts. I also owe David P Bird an apology.

I missed a qualifier in the third requirement for a JExocet.

3) The two target cells must be forced to reduce to different base digits. This is satisfied when all occurrences of a digit (solved or not) in the "S", cross-line, cells are contained by two houses. A way to check this is to consider how many lines would be needed to cover all of them.

Regards, Danny A. Jones

[Leren]

David P Bird wrote : 1) JE = true: two digits from (abc)

Hi David, I have no issues with your post except that it looks like you have a typo ; I think you meant JE = true: two digits from (abd)

If a digit has an S cell cover count > 2 (your c) then I try some "other" method to prove that if c is in the base cells then it must be in at least one target cell.

I use 2 methods to prove this : 1. Some simple in-band chains, or failing this, 2. A multi-digit expansion that assumes c is in each base cell and no target cell. If a contradiction is not found then the digit fails the test and so an Exocet is not found. If a contradiction is found then the digit passes the test.

Personally I prefer to use 1 before 2, not because 2 is any less logical than 1 but it makes me feel better, because I feel that I'm simulating what, in theory, a manual solver might be capable of. In fact, if no contradiction occurs in 2, then it may be that I have entirely solved the puzzle to find out that an Exocet does not exist - to me this would be an exercise in futility !

Interestingly I transposed the dobichev puzzle and found the Jexocet r1c7 r2c7 r5c8 r7c9 2789 with S cell counts a=2: - 1, b=7 - 2, c=7 - 2 and d=9 - 2. I used your single Truth theorem to eliminate 2 from the base and target cells and this solves the puzzle completely except for some basics, a grouped Skysrcaper/Finned fish and 2 ordinary Skysrapers. In that sense the other Exocets found in the non-transposed puzzle are ultimately extraneous.

While it's not my sub-thread I did not understand what blue was driving at in his post. There is no Exocet at r3c12,r1c5,r2c8 but there is one at r3c23,r1c5,r2c8 - maybe this was a typo, but I'll let you take that up with blue.

Leren

[Leren]

Just for the helluvit I wrote some code to detect digits that are forced into both target cells (and can therefore be eliminated from the base and target cells) along the lines of point 4 I mentioned a few posts back.

Applying this to the dobrichev puzzle the first Exocet I detected was r1c7 r2c7 r5c8 r9c9 2789 ; r8c9==r5c8; 9 is forced into both target cells => be removed from base and target cells.

This cracks the puzzle except for a Skyscraper or BUG+1. Yee haa (I suppose) !

While I was at it I applied the same technique to blue's puzzle and found an Exocet r3c2 r3c3 r1c5 r2c8 1234; Apparently 2 and 3 are forced into both target cells and can be removed from base and target cells.

This eventually completely solves the Exocet target cells and the whole puzzle. Is this good ?

Leren

[David P Bird]

I think you meant JE = true: two digits from (abd)"

No I did mean two from (abc). If (d) is true in the base cells its single 'S' cell must be true in in the base cell cross-line so it must occupy both targets. However it would be possible for (c) to be true in the base cells and just one target.

While it's not my sub-thread I did not understand what blue was driving at in his post. There is no Exocet at r3c12,r1c5,r2c8 but there is one at r3c23,r1c5,r2c8 - maybe this was a typo, but I'll let you take that up with blue.

The point behind Blues grid is that it satisfies the 'S' cell cover counts I wanted to explore to give an Almost JE which turns out to be false. The base cells hold (13) and the targets hold (1) and (4).

It shows my original conclusion when the JE is false was wrong. I've wasted some of my limited time chasing false trails but I have collected some potential derived inferences to check out which I'll report on later. Unfortunately they aren't as significant as the ones I thought I saw before.

DPB

[David P Bird]

This is my latest effort extensively revised after a helpful PM exchange with Blue:

When the number of cover houses needed for an (abcd) Almost JExocet are (a) =2, (b) =2, (c) >2, (d) =1
The alternatives are

1) JE = True: (d) will be false in the base and target cells.
2) JE = False: (c) will be true in the base cells and three 'S' cells.

So it's not such a win-win as I thought. However this does provide these derived inferences:

1) Weak link (d-ab#1) base cells ... which forces (c)base cells
2) Weak link (d)base cells – (all other digits)target cells ... if (d) is true in the base cells it must occupy both targets.
3) Weak link (x)base cells – (x)CellsSeeingBothTargets for x = (a) or (b) ... the digit must be true in at least one target
4)Weak link (ab)base cells – (d)Cell_inSameMini-Line ... explained below.
5) Strong link (c)base cells = (ab)JE ... when (c) is false then (d) must be false too and a (ab)JE will be true
6) Strong link (c)'S'Cell Swordfish = (abc)JE ... either three (c) 'S' cells will be true (making a Swordfish) or a (abc)JE will be

Inference 4): when the (ab)JE is true (d) must be true in two cross-line cells in the JE band in sight of the base cells. It therefore can't be true in the same mini-line as the base cells as then no second cell would be available.

Note that when (c) is true with either (a) or (b) one of the targets won't be constrained and will be free to contain any digit.

The list had to be revised because I suffered from a bout of brain rot and managed to mix up the different cases.

Code: Select all

 *--------------------------*--------------------------*--------------------------*
 | <5>     13489   123489   | 1236    12346 t 136      | 2368    <7>     1246     |
 | <6>     <7>     12348    | 1235    <9>     135      | 2358    12345 t 1245     |
 | 234 b   134 b   1234     | 123567  123467  <8>      | 2356    12345   <9>      |
 *--------------------------*--------------------------*--------------------------*
 | 2349    3459    <7>      | 1359    <8>     1359     | 259     <6>     125      |
 | <1>     3569    359      | 35679   367     <2>      | <4>     <8>     57       | ..3.
 | 29      5689    2589     | <4>     167     15679    | 2579    125     <3>      | 12..
 *--------------------------*--------------------------*--------------------------*
 | 379     1359    1359     | 123679  12367   <4>      | 23567   235     <8>      | 123.
 | 3479    <2>     3459     | 8       367     3679     | <1>     345     4567     | ..34
 | <8>     134     <6>      | 1237    <5>     137      | 237     <9>     247      |
 *--------------------------*--------------------------*--------------------------*
                   CLb                CL1                        Cl2

Almost JE (1234)r3c12, r1c5,r2c8 (cover houses (1)=2, (2)=2, (3)=3, (4)=1)
(3=12)r3c12 -[JE]- (12#1=35)r2c46 => r2c3,r3c45 <> 3 (r2c46 is in sight of both target cells)
(3=12)r3c12 -[JE]- (4)r1c5 = (47-6)r3c45 = (6)r3c7 => r3c7 <> 3 (r1c5 is a target cell)

These are the only unbranched AIC inferences I've found, but using forcing chains or net based methods much more can be achieved.
There is also a problem that concerns me but probably not many others. Because the JE is unproven if a derived inference from it is used further on in the chain this involves using memory and the chain ceases to be bidirectional. I can accept that approach only if the JE has been proved in a previous step because then it can be considered an established fact similar to a previous candidate elimination.

[Edits] the original posting was withdrawn and a revised version posted on 31st Jan 2015

[Leren]

Hi David, I've been staring at your last post for a while now and I'm afraid I just don't follow it. Rather than engage in a long winded description of what I think the problem is I'll just leave you with the following conundrum.

You say : (1)

Inference 4): when the (ab)JE is true (d) must be true in two cross-line cells in the JE band in sight of the base cells. It therefore can't be true in the same mini-line as the base cells as then no second cell would be available.

and further on : (2) (My underlining)

Almost JE (1234)r3c12, r1c5,r2c8 (cover houses (1)=2, (2)=2, (3)=3, (4)=1)
(3=12)r3c12 -[JE]- (12#1=35)r2c46 => r2c3,r3c45 <> 3 (r2c46 is in sight of both target cells)
(3=12)r3c12 -[JE]- (4)r1c5 = (47-6)r3c45 = (6)r3c7 => r3c7 <> 3 (r1c5 is a target cell)

Suppose the (ab) JE is True in your example puzzle, then r2c1 = 2 and r2c2 = 1. By (1) r2c3 can't be 4, and it can't be 1 or 2, so it must be 3 - but by (2) r2c3 <> 3

Leren

[David P Bird]

Suppose the (ab) JE is True in your example puzzle, then r3c1 = 2 and r3c2 = 1. By (1) r3c3 can't be 4, and it can't be 1 or 2, so it must be 3 - but by (2) r2c3 <> 3

Hi Leren,
You are a victim of your own typos! You're basing your arguments on eliminations in r2c3 but these should read r3c3. The earlier inference is then properly shown as (3)r2c3 so doesn't empty the cell as you've supposed.

I can't use that cell to extend the derived inferences from the pattern as in other cases it could contain non-base candidates.

DPB

[Leren]

Hi David, of course you are right, because of my typos there is no conflict with r2c3 <> 3 but there may be other inferences.

The various possibilities seem to boil down to 2 cases. A: either 3 is in the base r3c12 or B: 12 is in r3c12 and r3c3 = 3. So 3 is in one of r3c123. That would suggest - 3 r1c23, r2c3, r3c4578.

I've marked in red the eliminations not already seen by you. Does that look right ?

Leren

[David P Bird]

Leren,

As I was ploughing through the options it struck me that the pattern was a kraken node with three possible truth states:
1) A true JE with (3) false in the base cells
2) A true JE with (3) true in the base cells and one target
3) A false JE with (3) true in the base cells and false in both targets (one target may become unrestricted)

As you allow yourself far greater liberties in your methods than me you should this combination of cover houses fairly easy meat.

DPB

[Leren]

I now see that the extra eliminations I mentioned are only valid in the restricted case where the third cell in base cells mini-line only contains base digits.

Leren

[David P Bird]

Hi all,

I'm now struggling to find examples of double JExocets that have a true base digit common to both base sets (ie with three target cells) which I need to complete my descriptions of JE patterns. These are the three templates that seem to be theoretically possible for the true digits:

Code: Select all

*---------*---------*----------*  *---------*---------*----------*  *---------*---------*----------*
| AB AB c | .  /  . | .  .  .  |  | AB AB / | .  /  . | .  .  .  |  | AB AB / | .  /  . | .  .  .  |
| .  .  / | .  b  . | \  .  .  |  | .  .  c | .  b  . | \  .  .  |  | .  .  c | .  b  . | a  .  .  |
| .  .  . | .  \  . | a  BC BC |  | .  .  . | .  \  . | a  BC BC |  | .  .  . | .  \  . | \  BC BC |
*---------*---------*----------*  *---------*---------*----------*  *---------*---------*----------*
                         / = No base set1 digit        \ = No base set2 digit

Can anyone help please?

David PB

[daj95376]

Hi all,

I'm now struggling to find examples of double JExocets that have a true base digit common to both base sets (ie with three target cells) which I need to complete my descriptions of JE patterns. These are the three templates that seem to be theoretically possible for the true digits:

Code: Select all

*---------*---------*----------*  *---------*---------*----------*  *---------*---------*----------*
| AB AB c | .  /  . | .  .  .  |  | AB AB / | .  /  . | .  .  .  |  | AB AB / | .  /  . | .  .  .  |
| .  .  / | .  b  . | \  .  .  |  | .  .  c | .  b  . | \  .  .  |  | .  .  c | .  b  . | a  .  .  |
| .  .  . | .  \  . | a  BC BC |  | .  .  . | .  \  . | a  BC BC |  | .  .  . | .  \  . | \  BC BC |
*---------*---------*----------*  *---------*---------*----------*  *---------*---------*----------*
                         / = No base set1 digit        \ = No base set2 digit

Can anyone help please?

The following is based on my understanding of your request. Specifically with regard to the base cells.

Champagne has a file, "03 ED exo double.txt", that (supposedly) contains 279,043 double JExocet puzzles -- according to his definition of JExocet. I scanned this file for any of my QExocet patterns -- which should include JExocet patterns as a subset. Only three of the puzzles contain a double QExocet with three values in the four base cell's solutions. However, I doubt if any of the QExocets are JExocets.

Code: Select all

98.76.5..7..4.9....6....7...7..5.3....2.............1..3...6.87...8.36......7..5.

;1622942;GP;14_09;d;3;r7c1 r7c3 r8c5 r8c8 1249;r7c1 r7c3 r9c6 r9c7 1249;r8c8 r8c9 r9c2 r9c6 1249

 ### -1249- QExocet   Base = r8c89 <12>   Target = r7c1,r7c5
 ### -1249- QExocet   Base = r9c46 <29>   Target = r7c7,r7c1


98.76.5..7.4..........8.3..3.8...2...2...8.65..6.2.....5..3.6.......9..1.......5.

;1623876;GP;14_09;d;4;r4c8 r4c9 r6c2 r6c4 1479;r4c8 r4c9 r5c3 r5c5 1479;r6c4 r6c6 r4c2 r4c8 1479

 ### -1479- QExocet   Base = r4c89 <19>   Target = r6c4,r6c2
 ### -1479- QExocet   Base = r6c46 <17>   Target = r4c8,r4c2


98.76.5..7.4..........8.3..5.6.2....3.....2...2...8.65.5..3.6.......9..1.......5.

;1623896;GP;14_09;d;4;r4c4 r4c6 r5c2 r5c8 1479;r4c4 r4c6 r6c3 r6c7 1479;r5c8 r5c9 r4c2 r4c4 1479

 ### -1479- QExocet   Base = r4c46 <17>   Target = r5c8,r5c2
 ### -1479- QExocet   Base = r5c89 <19>   Target = r4c4,r4c2


In addition, champagne lists an exocet in each puzzle that I can't match. Consider the last puzzle listed above:

Code: Select all

98.76.5..7.4..........8.3..5.6.2....3.....2...2...8.65.5..3.6.......9..1.......5.

;1623896;GP;14_09;d;4
                     ;r4c4 r4c6 r5c2 r5c8 1479     equivalent to my first  QExocet
                     ;r4c4 r4c6 r6c3 r6c7 1479     ???
                     ;r5c8 r5c9 r4c2 r4c4 1479     equivalent to my second QExocet

 +--------------------------------------------------------------------------------+
 |  9       8       123     |  7       6       1234    |  5       124     24      |
 |  7       136     4       |  1259    159     1235    |  189     1289    2689    |
 |  126     16      5       |  1249    8       124     |  3       12479   24679   |
 |--------------------------+--------------------------+--------------------------|
 |  5       1479    6       |  149     2       147     |  14789   134789  34789   |
 |  3       1479    8       |  14569   14579   14567   |  2       1479    479     |
 |  14      2       179     |  3       1479    8       |  1479    6       5       |
 |--------------------------+--------------------------+--------------------------|
 |  1248    5       1279    |  1248    3       1247    |  6       24789   24789   |
 |  2468    3467    237     |  24568   457     9       |  478     23478   1       |
 |  12468   134679  12379   |  12468   147     12467   |  4789    5       234789  |
 +--------------------------------------------------------------------------------+
 # 162 eliminations remain

 ### -1479- QExocet   Base = r4c46 <17>   Target = r5c8,r5c2
 ### -1479- QExocet   Base = r5c89 <19>   Target = r4c4,r4c2


Champagne's second exocet, r4c4 r4c6 r6c3 r6c7 1479, seems questionable. Consider the following assignments:

Code: Select all

 r4c4=1   and   r1c6,r2c7,r3c2,r5c8,r6c1,r7c3,r9c5=1   and   (given) r8c9=1
 *--------------------------------------------------------------------------------*
 |  9       8       23      |  7       6      =1       |  5       24      24      |
 |  7       36      4       |  259     59      235     | =1       289     2689    |
 |  26     =1       5       |  249     8       24      |  3       2479    24679   |
 |--------------------------+--------------------------+--------------------------|
 |  5       479     6       | B149     2      B47      |  4789    34789   34789   |
 |  3       479     8       |  4569    4579    4567    |  2      =1       479     |
 | =1       2      T79      |  3       479     8       | T479     6       5       |
 |--------------------------+--------------------------+--------------------------|
 |  248     5      =1       |  248     3       247     |  6       24789   24789   |
 |  2468    3467    237     |  24568   457     9       |  478     23478 [=1]      |
 |  2468    34679   2379    |  2468   =1       2467    |  4789    5       234789  |
 *--------------------------------------------------------------------------------*


A perfectly acceptable template for base cell r4c4 being true for "1" and the target cells being false for "1".

_

[Edit: corrected JExocet reference to Exocet. This could explain why I didn't find DPB's pattern.]

[champagne]

Champagne's second exocet, r4c4 r4c6 r6c3 r6c7 1479, seems questionable. Consider the following assignments:

Code: Select all

 r4c4=1   and   r1c6,r2c7,r3c2,r5c8,r6c1,r7c3,r9c5=1   and   (given) r8c9=1
 *--------------------------------------------------------------------------------*
 |  9       8       23      |  7       6      =1       |  5       24      24      |
 |  7       36      4       |  259     59      235     | =1       289     2689    |
 |  26     =1       5       |  249     8       24      |  3       2479    24679   |
 |--------------------------+--------------------------+--------------------------|
 |  5       479     6       | B149     2      B47      |  4789    34789   34789   |
 |  3       479     8       |  4569    4579    4567    |  2      =1       479     |
 | =1       2      T79      |  3       479     8       | T479     6       5       |
 |--------------------------+--------------------------+--------------------------|
 |  248     5      =1       |  248     3       247     |  6       24789   24789   |
 |  2468    3467    237     |  24568   457     9       |  478     23478 [=1]      |
 |  2468    34679   2379    |  2468   =1       2467    |  4789    5       234789  |
 *--------------------------------------------------------------------------------*


A perfectly acceptable template for base cell r4c4 being true for "1" and the target cells being false for "1".

_

Hi Danny,

I am back in France, but not yet back home. You raise an interesting point, I'll comment on it ASAP (this could be a bug in my code).

[David P Bird]

DAJ, thanks for your search.

As you suspected, none of your finds complies with JExocet 'S' cell requirements.

The second and third are real curiosities as they can be considered Almost triple JEs which are true if (1)r2c7 is false. Considering the two cases then gives an elimination and if I resort to branching, it proves (1)r2c7 must be true, killing the JEs as (1) never occupies a JE target. There's some interesting truth balancing aspects though as in each row of the JE band there are two base cells and two targets to satisfy. We also know the 6 targets must contain at least three different digits to avoid a deadly pattern.

The targets you specified didn't make sense to me for a JE pattern however. At some point I would like some clarification about whether the target cells in a QExocet pattern must hold different base digits or not.

To make it clear, I'm currently looking for duplicate target cells for the two component patterns, not a base cell in one pattern that is a target cell for the other.

I have Champagnes 03 Double Exocet file and scanning some 200 puzzles by eye I didn't find one match.

I then went back to an earlier 'green' file and found 5 or 6 in the first 200 but they were all trivial. However it proved they should be included.
Next I tried a 'grey file' but it overwhelmed my text processors as it's enormous.

Today I managed to get a spreadsheet to load it (which took ages) which reports 889,598 records. Filtering those on a spreadsheet will be very laborious though. To get anything like a respectable speed I'll have to split out 500 records or so at a time.

DPB

[David P Bird]

Here's the analysis for the second of DAJ's puzzles

98.76.5..7.4..........8.3..5.6.2....3.....2...2...8.65.5..3.6.......9..1.......5.

Code: Select all

 *--------------------------*--------------------------*--------------------------*
 | <9>     <8>     123      | <7>     <6>     1234     | <5>     124     24       | -
 | <7>     136     <4>      | 1259    159     1235     | 189     1289    2689     | 1  9
 | 126     16      5        | 1249    <8>     124      | <3>     12479   24679    | 1
 *--------------------------*--------------------------*--------------------------*
 | <5>     1479 r  <6>      | 149 Q   <2>     147 Q    | 14789 p 134789  34789    |
 | <3>     1479 q  8        | 14569   14579 p 14567    | <2>     1479 R  479 R    |
 | 14 P    <2>     179 P    | 3       1479 r  <8>      | 1479 q  <6>     <5>      |
 *--------------------------*--------------------------*--------------------------*
 | 1248    <5>     1279     | 1248    <3>     1247     | <6>     24789   24789    | -
 | 2468    3467    237      | 24568   457     <9>      | 478     23478   <1>      |  47
 | 12468   134679  12379    | 12468   147     12467    | 4789    <5>     234789   | 1479
 *--------------------------*--------------------------*--------------------------*   
            CLP                       CLQ                CLR

Almost Triple JE: (1479) P:r6c13,r5c5,r4c7, Q:r4c46,r5c2,r6c7, R:r5c89,r4c2,r6c5 (digit(1) has three cover houses)

The base cells are annotated with capitals and their target cells with lower case letters.
Each base pair sees the targets for the other base pairs so one digit must be unique to each pair.
Digits (4), (7), & (9) must be true in at least one target cell for the base cells it occupies as they all have 2 cover houses.
It then follows that (1) must occupy one cell in each base pair and can be eliminated from the six target cells.
This reduces the puzzle to singles with none of the single JEs being true.